putting

$E=E_0\sin \omega t$ in the given equation we get,

$ Ri+\int\frac{i}{c}dt=E_0\sin\omega t$

Differentiating w.r.t. t we get,

$Ri+\int\frac{i}{c}dt=E_0\sin\omega t$

$ R\frac{di}{dt}+\frac{i}{c}=E_0\omega \cos\omega t $

Dividing throughout by R we get,

$\frac{di}{dt}+\frac{i}{Rc}=\frac{E_0}{R}\omega\cos\omega t $

This is a linear differential equation in i with $P=\frac{1}{RC}$ and $Q=\frac{E_0}{R}$

$I.F=e^{\int \frac{1}{RC}dt}=e^{\frac{t}{RC}} $

The solution is

$ie^{\frac{t}{RC}}=\int e^\frac{t}{RC}\frac{E_0}{R}\omega\cos\omega tdt+k$

$ie^{\frac{t}{RC}}=\frac{E_0}{R}\omega\int e^\frac{t}{RC}\cos\omega tdt+k $

$ ie^{\frac{t}{RC}}=\frac{E_0\omega}{R}\frac{e^\frac{t}{RC}}{\left( \frac{1}{RC}\right)^2+\omega^2}\left( \frac{1}{RC}\cos \omega t+\omega \sin \omega t\right)$

$ie^{\frac{t}{RC}}=\frac{E_0\omega}{R}\frac{e^{\frac{t}{RC}}}{\sqrt{\left( \frac{1}{RC}\right)^2+\omega^2}}\left( \frac{\frac{1}{RC}}{\sqrt{\left( \frac{1}{RC}\right)^2+\omega^2}}\cos\omega t + \frac{1}{\sqrt{\left( \frac{1}{RC}\right)^2+\omega^2}}\omega\sin\omega t\right)+k$

putting

$\frac{\frac{1}{RC}}{\sqrt{\left( \frac{1}{RC}\right)^2+\omega^2}} = \cos \phi\ and\ \frac{\omega}{\sqrt{\left( \frac{1}{RC}\right)^2+\omega^2}}=\sin \phi$

$ie^{\frac{t}{RC}}=\frac{E_0\omega}{R}\frac{e^{\frac{t}{RC}}}{\sqrt{\left( \frac{1}{RC}\right)^2+\omega^2}}\cos(\omega t-\phi)+k$

$\tan\phi=\frac{\omega}{\frac{1}{RC}}=RC\omega$

$ie^{\frac{t}{RC}}=\frac{E_0\omega Ce^{\frac{t}{RC}}}{\sqrt{R^2C^2\omega^2+1}}cos(\omega t-\phi)+k$

$i= \frac{E_0\omega C}{\sqrt{R^2C^2\omega^2+1}}cos(\omega t-\phi)+e^{\frac{-t}{RC}}k$